[初识行列式]行列式的计算

前一章整理了行列式的基本定义,接下来我们可以进行列式的计算.

1. 消零化基本型

• 适用条件:
  • 某行(列)已有足够多的0元素
  • 阶数不高

1. 例题1. 求n阶行列式${\left|\begin{array}{cccccc}a& b& 0& ...& 0& 0\\ 0& a& b& ...& 0& 0\\ 0& 0& a& ...& 0& 0\\ ...& ....& ....& ...& ...& ...\\ 0& 0& 0& ....& a& b\\ b& 0& 0& ....& 0& a\end{array}\right|}_{n\ast n}=?$
   解:按照第一列展开,有:
   $$\begin{gathered} D_n=\left|\begin{array}{cccccc}a& b& 0& ...& 0& 0\\ 0& a& b& ...& 0& 0\\ 0& 0& a& ...& 0& 0\\ ...& ....& ....& ...& ...& ...\\ 0& 0& 0& ....& a& b\\ b& 0& 0& ....& 0& a\end{array}\right|=a\left|\begin{array}{ccccc}a& b& ...& 0& 0\\ 0& a& ...& 0& 0\\ ....& ....& ...& ...& ...\\ 0& 0& ....& a& b\\ 0& 0& ....& 0& a\end{array}\right| \\ +(-1{)}^{n+1}b\left|\begin{array}{ccccc}b& 0& ...& 0& 0\\ a& b& ...& 0& 0\\ ....& ....& ...& ...& ...\\ 0& 0& ....& b& 0\\ 0& 0& ....& a& b\end{array}\right| \\ =a^n+(-1{)}^{n+1}{b}^n \end{gathered}$$

2. 爪形和异爪形行列式

• 爪形行列式:
  • 斜爪消平爪
• 异爪形行列式
  • n<=4时,直接展开
  • n>=4时,用递推法

2. 例题2.异爪形行列式:求$\left|\begin{array}{cccc}\lambda & -1& 0& 0\\ 0& \lambda & -1& 0\\ 0& 0& \lambda & -1\\ 4& 3& 2& \lambda +1\end{array}\right|=?$
   解:直接按照第四行展开有:
   $$\begin{gathered} D_n=\left|\begin{array}{cccc}\lambda & -1& 0& 0\\ 0& \lambda & -1& 0\\ 0& 0& \lambda & -1\\ 4& 3& 2& \lambda +1\end{array}\right|=4\ast (-1{)}^{1+4}\ast \left|\begin{array}{ccc}-1& 0& 0\\ \lambda & -1& 0\\ 0& \lambda & -1\end{array}\right|+3\ast (-1{)}^{2+4}\left|\begin{array}{ccc}\lambda & 0& 0\\ 0& -1& 0\\ 0& \lambda & -1\end{array}\right| \\ +2\ast (-1{)}^{3+4}\left|\begin{array}{ccc}\lambda & -1& 0\\ 0& \lambda & 0\\ 0& 0& -1\end{array}\right|+(\lambda +1)(-1{)}^{4+4}\left|\begin{array}{ccc}\lambda & -1& 0\\ 0& \lambda & -1\\ 0& 0& \lambda \end{array}\right| \\ =4+3\lambda +2{\lambda }^2+{\lambda }^3+{\lambda }^4 \end{gathered}$$
3. 例题3.爪形行列式:计算行列式$\left|\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 0& 0\\ 1& 0& 3& 0\\ 1& 0& 0& 4\end{array}\right|=?$
   解:用斜爪消平爪:$$D_n=\left|\begin{array}{cccc}1& 1& 1& 1\\ 1& 2& 0& 0\\ 1& 0& 3& 0\\ 1& 0& 0& 4\end{array}\right|=2\ast 3\ast 4\left|\begin{array}{cccc}1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}& 0& 0& 0\\ \frac{1}{2}& 1& 0& 0\\ \frac{1}{3}& 0& 1& 0\\ \frac{1}{4}& 0& 0& 1\end{array}\right|=-2$$

3. 拉普拉斯展开式

• 直接用定理:
  设A为m阶矩阵,B为n阶矩阵,有以下定理:
  $$\left|\begin{array}{cc}A& O\\ O& B\end{array}\right|=\left|\begin{array}{cc}A& C\\ O& B\end{array}\right|=\left|\begin{array}{cc}A& O\\ C& B\end{array}\right|=|A||B|$$
  $$\left|\begin{array}{cc}O& A\\ B& O\end{array}\right|=\left|\begin{array}{cc}C& A\\ B& O\end{array}\right|=\left|\begin{array}{cc}O& A\\ B& C\end{array}\right|=(-1{)}^{mn}|A||B|$$

4. 例题4:计算行列式$\left|\begin{array}{cccc}{a}_1& 0& 0& b_1\\ 0& a_2& b_2& 0\\ 0& b_3& a_3& 0\\ b_4& 0& 0& a_4\end{array}\right|$
   解:先将第二行和第四行互换,再将第二列和第四列互换,互换两次符号不变$$D_n=\left|\begin{array}{cccc}{a}_1& 0& 0& b_1\\ 0& a_2& b_2& 0\\ 0& b_3& a_3& 0\\ b_4& 0& 0& a_4\end{array}\right|=\left|\begin{array}{cccc}{a}_1& b_1& 0& 0\\ b_4& a_4& 0& 0\\ 0& 0& a_3& b_3\\ 0& 0& b_2& a_2\end{array}\right|=(a_1{a}_4-b_1{b}_4)(a_3{a}_2-b_2{b}_3)$$

4. 范德蒙行列式

• 直接用公式计算
  $$\left|\begin{array}{cccc}1& 1& ....& 1\\ x_1& x_2& ...& x_n\\ x_1^2& x_2^2& .....& x_n^2\\ ...& ....& ...& ....\\ x_1^{n-1}& x_2^{n-1}& ......& x_n^{n-1}\end{array}\right|=\prod _{1\le i<j\le n}(x_j-x_i)$$

理解:${\prod }_{1\le i<j\le n}(x_j-x_i)$:高年级欺负低年级,并且所有的高年级都要欺负到所有的低年级

5. 例题5:计算行列式:$\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ b+c& a+c& a+b\end{array}\right|$
   解:第一眼看题目应该会想到范德蒙行列式,但是没有全为1的一行,所以我们需要自己构造,将第一行加到第三行上去,并提出公因子(a+b+c)即能得到想要的结果:
   $$\begin{gathered} D_n=\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ b+c& a+c& a+b\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right| \\ =(-1)(-1)(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=(a+b+c)(c-b)(c-a)(b-a) \end{gathered}$$

5. 数学归纳法和递推法

• 当涉及n阶行列式时候,可以考虑用数学归纳法

6. 例题6:计算$\left|\begin{array}{cccccc}2& -1& 0& ...& 0& 0\\ -1& 2& -1& ...& 0& 0\\ 0& -1& 2& ...& 0& 0\\ ...& ...& ...& ....& ....& ....\\ 0& 0& 0& ...& 2& -1\\ 0& 0& 0& .....& -1& 2\end{array}\right|$
   解:将Dn中的第2,3…n列加到第1列上去,并按照第1列展开:
   $$\begin{gathered} D_n=\left|\begin{array}{cccccc}1& -1& 0& ...& 0& 0\\ 0& 2& -1& ...& 0& 0\\ 0& -1& 2& ...& 0& 0\\ ...& ...& ...& ....& ....& ....\\ 0& 0& 0& ...& 2& -1\\ 1& 0& 0& .....& -1& 2\end{array}\right|=D_{n-1}+1(-1{)}^{n+1}\left|\begin{array}{ccccc}-1& 0& ...& 0& 0\\ -2& -1& ...& 0& 0\\ -1& 2& ...& 0& 0\\ ...& ...& ....& ....& ....\\ 0& 0& ...& -1& 0\\ 0& 0& ...& 2& -1\end{array}\right| \\ =D_{n-1}+(-1{)}^{n+1}(-1{)}^{n-1} \\ =D_{n-1}+1 \end{gathered}$$
   根据递推关系式有$D_n=D_1+(n-1)=2+n-1=n+1$

7. 例题7,计算行列式:$\left|\begin{array}{cccccc}{a}_1& -1& 0& .....& 0& 0\\ a_2& x& -1& ...& 0& 0\\ a_3& 0& x& .....& 0& 0\\ ...& ....& ...& ......& .....& ....\\ a_{n-1}& 0& 0& ......& x& -1\\ a_n& 0& 0& .......& 0& x\end{array}\right|$
   解:此题需要注意只能从最后一行展开,否则无法找出递推公式:
   $$\begin{gathered} D_n=(-1{)}^{n+1}{a}_n\left|\begin{array}{ccccc}-1& 0& .....& 0& 0\\ x& -1& ...& 0& 0\\ 0& x& .....& 0& 0\\ ....& ...& ......& .....& ....\\ 0& 0& ......& x& -1\end{array}\right|+x(-1{)}^{n+n}{D}_{n-1} \\ =a_n+x{D}_{n-1} \end{gathered}$$
   由递推公式可得$D_n=a_n+a_{n-1}x+a_{n-2}{x}^2+.......+a_2{x}^{n-2}+a_1{x}^{n-1}$